poj 2334 Simple prefix compression(胡搞)

最新推荐文章于 2020-04-21 08:10:49 发布

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本文介绍了一种数据库中使用的字符字段压缩技术——前缀压缩。该技术通过比较相邻字符串的公共前缀来减少存储空间。文章提供了一个示例程序，用于演示如何计算一组字符串经前缀压缩后的最小总长度。

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Simple prefix compression

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 1861		Accepted: 813

Description

Many databases store the data in the character fields (and especially indices) using prefix compression. This technique compresses a sequence of strings A1, ..., A_N by the following method: if there are strings Ai = a_i,1a_i,2...a_i,p and A_{i + 1} = a_i+1,1a_i+1,2...a_i+1,q
such that for some j ≤ min(p, q) a_i,1 = a_i+1,1, a_i,2 = a_i+1,2, ... a_i,j = a_i+1,j, then the second string is stored as [j]a_i+1,j+1a_i+1,j+2... a_i+1,q, where [j] is a single character with code j.

If j = 0, that is, strings do not have any common prefix, then the second string is prefixed with zero byte, and so the total length actually increases.

Constraints
1 ≤ N ≤ 10000, 1 ≤ length(Ai) ≤ 255.

Input

First line of input contains integer number N, with following N lines containing strings A1 ... A_N

Output

Output must contain a single integer -- minimal total length of compressed strings.

Sample Input

3
abc
atest
atext

Sample Output

tips：马德，至今未看懂题意；看着别人的题解慢慢的胡搞；

具体来说每一次都统计与前面一个字符串的前缀数量；然后将长度-前缀数量的长度累加起来；然后胡搞没看懂题

#include<iostream>
#include<cstring>
#include<string>


using namespace std;

int n;
string last,now; 
int main()
{
	cin>>n;int ans=0;
	while(n--)
	{
		last=now;
		cin>>now;
		
		int sz=0;
		for(int i=0;i<last.size()&&i<now.size();i++)
		{
			if(now[i]!=last[i])break;
			sz++;//统计前缀长度 
		}
		ans+=(now.size()-sz+1);//非前缀长度； 
		
	}
	cout<<ans-1<<endl; 
	return 0;
 }